Question

Use the appropriate formula to determine the most probable speed (in miles per hour) for N2 molecules at 3.00 x 102 K. (As you do this, be sure to show your units explicitly.) Repeat the calculation for N2 molecules at 1.00 x 103 K.

Answer 1

Use the relation (derivation can be obtained in texts on kinetic theory of gases)

V_mp = √2*R*T/M where v_mp is the most probable speed of the gas molecules; R is the gas constant, T = absolute temperature of the gas and M = molecular weight of the gas.

Use the value of R = 8.314 kg m² s^-2mol^-1 K^-1; T = 3.00*10² K and M = M.W. of N₂ expressed in kg/mol = (2*14.0067) g mol^-1 = 28.0134 g mol^-1 = (28.0134 g mol^-1)*(1 kg/1000 g) = 0.0280134 kg mol^-1. Plug in values and obtain

V_mp = √2*(8.314 kg m² s^-2 mol^-1 K^-1)*(3.00*10² K)/(0.0280134 kg mol^-1) = √(178071.9227 m² s^-2) = 421.98569 m s^-1 = (421.98569 m)/(1 s) = (421.98569 m)*(1 mile/1609.34 m)/[(1 s)*(1 min/60 s)*(1 h/60 min)] = (0.262210 mile)/(0.0002778 h) = 943.8804 mile/h ≈ 943.88 mile/h (ans).

Here T = 1.00*10³ K; therefore,

V_mp = √2*(8.314 kg m² s^-2 mol^-1 K^-1)*(1.00*10³ K)/(0.0280134 kg mol^-1) = 770.436938 m/s = (770.43698 m)/(1 s) = (770.43698 m)*(1 mile/1609.34 m)/[(1 s)*(1 h/3600 s)] = (0.478728 mile)/(0.0002778 h) = 1723.2829 mile/h ≈ 1723.28 mile/h (ans).